Question: Let $x$ and $y$ be positive real numbers.  Find the maximum value of
\[\frac{(x + y)^2}{x^2 + y^2}.\]
Answer: We claim that the maximum value is 2.  Note that for $x = y,$
\[\frac{(x + y)^2}{x^2 + y^2} = \frac{4x^2}{2x^2} = 2.\]The inequality $\frac{(x + y)^2}{x^2 + y^2} \le 2$ is equivalent to
\[(x + y)^2 \le 2x^2 + 2y^2,\]which in turn simplifies to $x^2 - 2xy + y^2 \ge 0.$  We can write this as $(x - y)^2 \ge 0.$  This inequality holds, and since all our steps are reversible, the inequality $\frac{(x + y)^2}{x^2 + y^2} \le 2$ also holds.  Hence, the maximum value is $\boxed{2}.$